The equation of a circle $C$ is $x^2+y^2-2x-18y+78 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-2x) + (y^2-18y) = -78$ $(x^2-2x+1) + (y^2-18y+81) = -78 + 1 + 81$ $(x-1)^{2} + (y-9)^{2} = 4 = 2^2$ Thus, $(h, k) = (1, 9)$ and $r = 2$.